Proving [\hat{x}, \hat{k}]=i in Operator Notation

[latentcorpse]

Ok so i need to prove that [ \hat{x} , \hat{k} ] = i in this operator notation where
\hat{x}=\int dx | x \rangle \langle x | x
\hat{k}=\int dk | k \rangle \langle k | k
and \langle x | k \rangle = \frac{1}{\sqrt{2 \pi}}e^{ikx}

so i have worked out
\hat{x} \hat{k} = \int dx \hat{k} | x \rangle \langle x | x
but \hat{k} | x \rangle = i \frac{\partial}{\partial x} | x \rangle
so \hat{x} \hat{k} = \int dx i \frac{\partial x}{\partial x} | x \rangle \langle x|
=i \int dx | x \rangle \langle x | = i \hat{1}=i where \hat{1} is the unit/identity operator

similarly i find that \hat{k} \hat{x}=-i
and so [\hat{x},\hat{k}]=i-(-i)=2i
i can't see where i was supposed to get rid of that factor of 2 though?

thanks.

 

[thebigstar25]

i just have one comment (in line 8), why did you assume that the partial operator will just be applied on x? I am curious since I am taking the same course now ..

 

[latentcorpse]

i think it's because the bra and the ket are basis states so their derivative would be zero. It says on this article
http://en.wikipedia.org/wiki/Matrix_mechanics
that
D \int \psi(x) | x \rangle dx = \int \psi'(x) | x \rangle dx
where D is a derivative operator.

 

[thebigstar25]

one other question .. What is the effect if we are applying x operator on k state? .. I want to try working out this problem maybe i would be helpful ..

 

[latentcorpse]

it's as follows:

\hat{x}|k \rangle = \int x | x \rangle \langle x | k \rangle dx = \int \frac{x}{\sqrt{2 \pi}} e^{-ikx} | x \rangle dx = \frac{dx}{\sqrt{2 \pi}} i \frac{\partial}{\partial k} e^{-ikx} | x \rangle = \int dx i \frac{\partial}{\partial k} |x \rangle \langle x | k \rangle = i \frac{\partial}{\partial k} | k \rangle

 

[vela]

\hat{x} \hat{k} = \int dx\, \hat{k} | x \rangle \langle x | x

You seem to have mistakenly commuted the operators. The order of x-hat and k-hat are different on the LHS and RHS.

 

[latentcorpse]

You seem to have mistakenly commuted the operators. The order of x-hat and k-hat are different on the LHS and RHS.

so the one I've actually shown the working for is \hat{k} \hat{x}. This gives me the i that i want so i just have to show \hat{x} \hat{k}=0

i have \hat{x} \hat{k} = \int dx x | x \rangle \langle x | \hat{k}

i don't know what to do now. should i sub for the k operator? then i'll have an integral over both x and k and i'll still have something operating to the right when there's nothing on the right to operate on.

 

[vela]

so \hat{k} \hat{x} = \int dx \, i \frac{\partial x}{\partial x} | x \rangle \langle x|x

This isn't quite right either (even after correcting the order of x and k). It might be clearer to see what's going on if you consider

\hat{k} \hat{x}|\psi(x)\rangle = \int dx \, i \frac{\partial}{\partial x} | x \rangle \langle x|x|\psi(x)\rangle

 

[latentcorpse]

This isn't quite right either (even after correcting the order of x and k). It might be clearer to see what's going on if you consider

\hat{k} \hat{x}|\psi(x)\rangle = \int dx \, i \frac{\partial}{\partial x} | x \rangle \langle x|x|\psi(x)\rangle

not following this. how did you know to change \hat{k} to i \frac{\partial}{\partial x} without doing the lines i did above to prove it?

anyway i get following it through
\int dx i \partial_x | x \rangle \langle x | \hat{x} | \psi \rangle
=i \int dx \int dx' \partial_x |x \rangle \langle x | x' | x' \rangle \langle x' | \psi \rangle
=i \int dx \partial_x | x \rangle \langle x | x \psi(x)
=i \partial_x (x \psi(x)) = i ( \psi(x) + x \psi'(x))
which looks a bit...erm...off!

 

[vela]

not following this. how did you know to change \hat{k} to i \frac{\partial}{\partial x} without doing the lines i did above to prove it?

I was just picking up from the middle of what you did. You'd still want to show the steps you did earlier to get to this point.

anyway i get following it through
\int dx i \partial_x | x \rangle \langle x | \hat{x} | \psi \rangle
=i \int dx \int dx' \partial_x |x \rangle \langle x | x' | x' \rangle \langle x' | \psi \rangle
=i \int dx \partial_x | x \rangle \langle x | x \psi(x)
=i \partial_x (x \psi(x)) = i ( \psi(x) + x \psi'(x))
which looks a bit...erm...off!

Actually, this is correct. So you can see that \hat{k}\hat{x} is the operator represented by i(1+x\partial_x), and if you look at that last term, you should see it's the representation of \hat{x}\hat{k}. Hence, the commutator is...?

 

[latentcorpse]

I was just picking up from the middle of what you did. You'd still want to show the steps you did earlier to get to this point.

Actually, this is correct. So you can see that \hat{k}\hat{x} is the operator represented by i(1+x\partial_x), and if you look at that last term, you should see it's the representation of \hat{x}\hat{k}. Hence, the commutator is...?

well it will be i.
this however, relies on me being able to show that \hat{x} \hat{k} = x \partial_x

\hat{x} \hat{k} | \psi \rangle = \int dx x | x \rangle \langle x| \hat{k} \psi \rangle
= \int dx \int dk x | x \rangle \langle | k \rangle \langle k | \psi \rangle k
=\int dx \int dk x | x \rangle (-i \partial_x) \langle x | k \rangle \langle k | \psi \rangle
=-i \int dx x | x \rangle \partial_x \langle x | \psi \rangle
=-i x \partial_x | \psi \rangle

which cancels the other term leaving just the i

also the term in the previous post should have been in terms of the ket |psi \rangle instead of \psi(x) if we're being pedantic about the algebra, shouldn't it?

 

[vela]

also the term in the previous post should have been in terms of the ket |psi \rangle instead of \psi(x) if we're being pedantic about the algebra, shouldn't it?

Yes, you're right. I was just looking at the derivative part. You got rid of one of the integrals when you shouldn't have. You had:

\hat{k}\hat{x}|\psi\rangle = i \int dx \int dx'\, \partial_x |x\rangle\langle x| x' | x'\rangle \langle x'|\psi\rangle = i \int dx \int dx' \, \partial_x |x\rangle \langle x|x' \rangle x'\psi(x')

You can then do one integral because of the <x|x'> delta function to get:

\hat{k}\hat{x}|\psi\rangle = i \int dx \, |x\rangle \partial_x (x\psi(x))

P.S. It looks like there might be a sign error somewhere. I'm not sure what the proper signs are, but I figure you can straighten all that out.

 

[thebigstar25]

=-i \int dx x | x \rangle \partial_x \langle x | \psi \rangle
=-i x \partial_x | \psi \rangle

im just wondering, in the last step where you carried on the integral, is it really fine to ignore the fact you have an operator on the state x, and take out the identity operator?? .. I just have a feeling that there is something missing here? ..

 

[vela]

No, it's not. You should end up with

-i \int dx\,|x\rangle x\partial_x\psi(x)

 

[latentcorpse]

No, it's not. You should end up with

-i \int dx\,|x\rangle x\partial_x\psi(x)

surely the final answer in post 9 should have an integral in it as well then?

 

[vela]

Yes, that's what I said in post 12.